1c is false.
f(x)=x2 => f'(x)=2x
2x=2 at x=1 which is not an elemnt of (1,3).
Let f be twice diff function satisfying f(1)=1,f(2)=4,f(3)=9 then:
a)f"(x)=2
b)f'(x)=2
c)there exists at least one x E (1,3) such that f ' (x)=2
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10 Answers
1
341You cannot assume f(x) = x2.
Instead look at the function g(x) = f(x) - x2
We have g(1) = g(2) = g(3) = 0
What conclusions can now be drawn?
21BTW wat is the questn heree???
do v hav to findthe functtn f ??
62no you dont tapan..
you have to just see which of these options are possible.
21oh ok!!
then using prophet suir's g(x) we can easily say
options b & c are wrong, wat is left shud be the ans
24but ans is given to be c........................[7][7][7]
HELP.................
341in that case the question should have been there exists x ε [1,3] such that f"(x) = 2
24maybe sir thats misprint..............but it is 2005 question.....
341In that case the question is wrong. The options should be satisfied by any function satisfying the given conditions. And as sky has pointed out f(x) = x2 does not satisfy the three options.
Better check your source